2
$\begingroup$

Prove that the limit $$\lim_{x\to1}\frac{\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du}{x-1}$$ exists and equals 4.

My thoughts involving this limit are that I will have to use L'hospital's rule at some point as the denominator will equal zero if I substitute 1 in for x, but I don't know how to handle the integral in the numerator.

$\endgroup$
3
  • $\begingroup$ At first, I wasn't sure how to approach it through the FTC because I wasn't sure how to show the numerator was zero, but I just now see how the integral from 2 to 2 is simply zero. This is something for me to go off of here. $\endgroup$ – Peetrius Jul 24 '18 at 11:04
  • $\begingroup$ I think the limit is not 1... $\endgroup$ – Szeto Jul 24 '18 at 12:22
  • $\begingroup$ Thanks for the heads up. I fixed it. $\endgroup$ – Peetrius Jul 25 '18 at 1:12
2
$\begingroup$

Let $f(x):=\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du$. Then

$$\lim_{x\to1}\frac{\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du}{x-1}= \lim_{x\to1}\frac{f(x)-f(1)}{x-1}=f'(1).$$

Can you proceed ?

$\endgroup$
1
  • $\begingroup$ I think I see where this is going. So the structure of the limit at hand looks like the limit definition of the derivative, so I just need to take the derivative of $f(x)$ and then plug in 1, and I've evaluated the limit? $\endgroup$ – Peetrius Jul 24 '18 at 11:10
2
$\begingroup$

Since the integrand $f$ is continuous by mean value theorem the integral in numerator is equal to $2(x-1)f(c)$ for some $c$ between $2$ and $2x$. Thus the expression under limit is equal to $2f(c)$ and as $x\to 1$ we have $c\to 2$ so that by continuity of $f$ we have the desired limit as $2f(2)=4$.

$\endgroup$
1
$\begingroup$

Hint: $$\dfrac{d}{dx}\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du=\dfrac{d}{dx}(2x) \left(\frac{2x}{1+\ln(2x/2)}\right) - \dfrac{d}{dx}(2) \left(\frac{2}{1+\ln(2/2)}\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.